Question 149469
They all have a solution. Do you mean if they have a real solution?


To find out if they have real solutions or not, you can use the discriminant formula.


1)



From {{{2x^2-10x+25}}} we can see that {{{a=2}}}, {{{b=-10}}}, and {{{c=25}}}



{{{D=b^2-4ac}}} Start with the discriminant formula



{{{D=(-10)^2-4(2)(25)}}} Plug in {{{a=2}}}, {{{b=-10}}}, and {{{c=25}}}



{{{D=100-4(2)(25)}}} Square {{{-10}}} to get {{{100}}}



{{{D=100-200}}} Multiply {{{4(2)(25)}}} to get {{{(8)(25)=200}}}



{{{D=-100}}} Subtract {{{200}}} from {{{100}}} to get {{{-100}}}



Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.



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From {{{2x^2-6x+5}}} we can see that {{{a=2}}}, {{{b=-6}}}, and {{{c=5}}}



{{{D=b^2-4ac}}} Start with the discriminant formula



{{{D=(-6)^2-4(2)(5)}}} Plug in {{{a=2}}}, {{{b=-6}}}, and {{{c=5}}}



{{{D=36-4(2)(5)}}} Square {{{-6}}} to get {{{36}}}



{{{D=36-40}}} Multiply {{{4(2)(5)}}} to get {{{(8)(5)=40}}}



{{{D=-4}}} Subtract {{{40}}} from {{{36}}} to get {{{-4}}}



Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.



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3)



From {{{s^2-4s+4}}} we can see that {{{a=1}}}, {{{b=-4}}}, and {{{c=4}}}



{{{D=b^2-4ac}}} Start with the discriminant formula



{{{D=(-4)^2-4(1)(4)}}} Plug in {{{a=1}}}, {{{b=-4}}}, and {{{c=4}}}



{{{D=16-4(1)(4)}}} Square {{{-4}}} to get {{{16}}}



{{{D=16-16}}} Multiply {{{4(1)(4)}}} to get {{{(4)(4)=16}}}



{{{D=0}}} Subtract {{{16}}} from {{{16}}} to get {{{0}}}



Since the discriminant is equal to zero, this means that there is one real solution